Match List I with List II.

	List I (Molecule)		List II (Number and types of bond/s between two carbon atoms)
A.	ethane	I.	one σ-bond and two π-bonds
B.	ethene	II.	two π-bonds
C.	carbon molecule, C2	III.	one σ-bond
D.	ethyne	IV.	one σ-bond and one π-bond

Choose the correct answer from the options given below:

The correct answer is A-III, B-IV, C-II, D-I.

Let us analyze the nature of the bonding between the two carbon atoms in each of the given molecules by looking at their structures and molecular orbital configurations:

A. Ethane ($C_2{H}_6$):
In ethane, each carbon atom is $s{p}^3$ hybridized. The bond between the two carbon atoms ($C-C$) is formed by the axial overlap of two $s{p}^3$ hybrid orbitals, which results in a single σ-bond (sigma bond). Therefore, A matches with III.

B. Ethene ($C_2{H}_4$):
In ethene, each carbon atom is $s{p}^2$ hybridized. The double bond between the two carbon atoms ($C=C$) consists of one σ-bond (formed by head-on $s{p}^2-s{p}^2$ overlap) and one π-bond (formed by lateral overlap of unhybridized $2p$ orbitals). Therefore, B matches with IV.

C. Carbon molecule ($C_2$):
According to Molecular Orbital (MO) Theory, the electronic configuration of $C_2$ (having 12 electrons) is:
$\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^2{\sigma }^{*}2s^2(\pi 2p_x^2=\pi 2p_y^2)$
The last four valence electrons occupy the bonding $\pi 2p_x$ and $\pi 2p_y$ molecular orbitals. As a result, the double bond in the $C_2$ molecule is made up of two π-bonds, with no σ-bond contributing to the bond order of 2. Therefore, C matches with II.

D. Ethyne ($C_2{H}_2$):
In ethyne, each carbon atom is $sp$ hybridized. The triple bond between the two carbon atoms ($C\equiv C$) consists of one σ-bond (formed by head-on $sp-sp$ overlap) and two π-bonds (formed by lateral overlaps of the two pairs of unhybridized $2p$ orbitals). Therefore, D matches with I.

Combining all the matches, we get:
A-III, B-IV, C-II, D-I